Prove there is a subspace of $V$ isomorphic to $T(V)$ 
If $T:V\to V$ is a linear transformation and $T(V)$ is of finite dimension then prove that there is a subspace $U$ of $V$ isomorphic to $T(V)$ and then show that, if $x,y\in V$, then $(x+U)\cap (y+Ker(T))$ consists of only one vector.

If dim($V$) is finite then since dim$(V)\ge$ dim$(T(V))$ from Rank Nullity theorem, we can choose subset $U$ of basis of $V$ whose dim equals $T(V)$ then Span$(U)$ will be isomorphic to $T(V))$, but what to do in infinte case or any other proof which covers both cases at the same  time? 
 A: To deal with all dimension possibilities at once, write $V = \ker T \oplus U$ for $U$ a complementary subspace to $\ker T$; this can be done even for $V$ infinite-dimensional (see this MSE post ).
Now argue that $T: \ker T \oplus U \rightarrow V$ restricts to an isomorphism of $U$ with $T(V)$. To do this, you might use the first isomorphism theorem for vector spaces; or simply argue that $T|_U$ is injective and surjective onto $T(V)$, using properties of the direct sum decomposition.
Given that $T$ restricts to an isomorphism of $U$ with $T(V)$, we can verify that 
$$(\forall x,y \in V) \,\, \,\,(x+U)\cap (y+\ker T) \text{ is a singleton }. \,\,\, (*)$$ Let $x,y$ be given. If $x+u=y+s$ for $u \in U$, $s \in \ker T$, then: $$T(y-x)=T(u-s)=T(u)-T(s)=T(u),\,\,\,\,$$ where the last equality follows from the fact that $s \in\ker T$. Since $T|_U$ is an isomorphism, there is a unique element $u \in U$ so that $T(u)=T(x-y)$. Then $s=x+u-y \in \ker T$ is the unique element of the kernel so that $x+u=y+s$.
Note that $(*)$ forces you to consider a "natural" construction, such as the one above. It is true that any subspace $U \subset V$ with the same dimension as $T(V)$ is isomorphic to $T(V)$. But the isomorphism between $U$ and $T(V)$ will in general be arbitrary, and in the proof of $(*)$ we used that the isomorphism was given by a specific map (the restriction of $T$ to $U$). 
A: Let $\{w_1,w_2,\dots,w_k\}$ be a basis of $T(V)$ and choose $v_i\in V$ such that $T(v_i)=w_i$ $(i=1,2,\dots,k)$. It's easy to see that $\{v_1,v_2,\dots,v_k\}$ is linearly independent. If $U=\langle v_1,v_2,\dots,v_k\rangle$ is the subspace spanned by these vectors, then the restriction of $T$ to $U$ is an isomorphism between $U$ and $T(V)$.
Note that there's no assumption on the dimension of $V$.
Now, if $z\in(x+U)\cap(y+\ker T)$, then $z=x+u=y+v$ for some $u\in U$ and $v\in\ker T$. Thus $u=y-x+v\in U$ and $T(u)=T(y-x)$. If $z'=x+u'=y+v'$ has the same property as $z$, then $T(u')=T(y-x)=T(u)$, so $u=u'$ and $z'=z$.
